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a+2d in Java Print qrcode in Java a+2d




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a+2d using jsp torender qr-code in asp.net web,windows application Android a +d 2 So (a+d) =- a(a+2d),giving a 2 +2ad+d 2 =a 2 +2ad andhence d =0. Since ar =a+d =a,r =1. The only arithmetic sequences which are also geometric are constant, of the form a, a, a, .

.. for some number a.

. 31. (a) $1000 swing qr-codes (1.15)2 = $1322.

50. (b) We seek t such that 1000(. 1 5)t = 2000.

Thus, 1.15t = 2, t = log 2/ log 1.15.

4.9595 years. = 0.. = 1 -1 as fk- 1 = fk+l- fk, we see that fo = f2 -f Rewriting fk+l = fk + fk f-5 = 5, f-6 = -8. In general, f-= (- l)"+1fn f-2 =-1, f-3 = 2, f-4 =-3,. Similarly, f-l = 1,. The Fibonacci sequence is the sequence 1, 1, 2, 3, 5, 8...

, each term after the first two being the sum of the 2 f= . So the assertion is correct for n = 1. Let k > I two previous terms.

For n = 1, f2fl = I(]) = 1= 7. We wish to prove the and assume the assertion is correct for n = k; that is, assume that fk+lfk = Tk. 2 f But fk+2 = fk+1 + fk, SO, assertion is correct for n = k + I; that is, we wish to prove that fk+2fk-l =.

fk 2fk 1 = (f applet QR Code 2d barcode k+1 + fk)fk 1 = f7+1 + fk~lfk = fk+] +>j=1 fi (using the induction hypothesis) = ,+ desired. By the Principle of Mathematical Induction, we conclude that the assertion is correct for all n > 1..

t2 as The induction applet QR Code hypothesis is valid only for n in the interval 3 < n < k because all integers in this problem are at least as large as no = 3. But the induction hypothesis is applied to the integer k -2; this is not valid if k = 4..

42. (a) aI = l, a2 = 2 (2 and l + 1), a3 = 4, as given. We can write 4 = 4 = 3 + I = I + 3 = 2 + I + I = I + 2 + I + 1 + I + so a4 = 8.

Careful counting shows that a 5 = 16. =I 1 + + 2 =2 +2 44. (a) an.

2a, 1. The se QR-Code for Java quence is the sequence 2, 4, 8, ..

.. of powers of 2. 47. (b) fl(l) Exercises 5.3 1. a,=. f(l). [4(1) -l]/3 = 1. f (1) =. f(l) = f(f(l)). f(l). 1. This continues: fn(l) = I for all n. a, =- 5 (3f) + 6n (3) an, = (-4 + 15)n (-4 -15). -2(-1)" + 2(3n) =2[(-I)n+l + 3 n (b) an = 2(3n) + 10. (a). an = 2(3f) + 2(-5)f;. =4";. (-5)f -2.. 13. (a) an (b) a, = 2(8n) - 4".. = 2 4. (c) For n = 0, 2(80) -40. = 1, as required. 2 (8. Let k > 0 and assume that ak k) -. 4 k.. We wish to prove that ak+1 - 4k] + 2 (8. 8 k+1 2 (8. k+l) -. 4 k+1. ak+l = ak + 8 k+1 4 [ 2 ( 8 k). (by the induction hypothesis). = 8k+1 - 4k+1 + 8 k+1 k+1) -. 4 k+1 as desired. B spring framework qr bidimensional barcode y the Principle of Mathematical Induction, the result is true for all n > 0. 16.

(a) an = 4f + 3; (b) a, = 10(4) -(3n + 6)2n.. The nth term of the Fibonacci sequence is a,_1 =. 1(-1)nV I/-( 2). 2)n _ I (1-,(. SO la,_1. Thus, lan-, < Now II(-V5)nl 2) o /5(. 5 1 < 1. 2 since I1-V51 2 I+V5)"I 7i ( 22 5. 2. Remembering that S-28. Solutions to Selected Exercises a,-, is an in teger, the result follows from the fact that there is precisely one integer within number. of any given real. 22. (a) Since x is a root of the characteristic polynomial, x 2 cxn-2(rX + S) = Cx,2. rx + s. Hence jar Denso QR Bar Code , ran_1 + sa,-2 = r(cx"-1) + s(cx"-2). (X ) = cXn = an. (b) We know t QR Code for Java hat pn = rpn-l + SPn-2 and qn = rqn-l + sqn-2. Thus, Pn + qn = rPnt + SPn-2 + rqn-i + sqn-2 = r(pn-1 + qn-]) + s(pn-2 + qn-2). (c) If xl and x2 are the characteristic roots, part (a) tells us that clx" and c2x" both satisfy the recurrence relation while part (b) says that c1xln + C2X2" is also a solution.

If either xl or x2 is 0, one initial condition determines the single unknown constant. Otherwise, two initial conditions will determine cl and c2 for they determine two linear equations in the unknowns Cl, C2 which, because xl :A X2, must have a unique solution. 23.

(a) The characteristic polynomial is x2-rx-s with characteristic roots x = ri2.
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